Explanation

(d)
$\lim_{x \rightarrow \alpha}\frac{1 - \cos{(ax^{2} + bx + c)}}{(x - \alpha)^{2}} = \lim_{x \rightarrow \alpha}\frac{2\sin^{2}\left( \frac{ax^{2} + bx + c}{2} \right)}{(x - \alpha)^{2}}$
[Since $\alpha$ and $\beta$ are the roots of $ax^{2} + bx + c = 0$, so it can be written as $a(x - \alpha)(x - \beta) = 0\rbrack$
$= \lim_{x \rightarrow \alpha}\frac{2\sin^{2}\left( \frac{a(x - \alpha)(x - \beta)}{2} \right)}{(x - \alpha)^{2}}$
$= \lim_{x \rightarrow \alpha}\frac{2\sin^{2}\left( \frac{a}{2}(x - \alpha)(x - \beta) \right)\left( \frac{a}{2} \right)^{2}(x - \beta)^{2}}{\left\lbrack \left( \frac{a}{2} \right)(x - \alpha)(x - \beta) \right\rbrack^{2}}$
$= \lim_{x \rightarrow \alpha}\ 2\left( \frac{a}{2} \right)^{2}(x - \beta)^{2} = \frac{a^{2}}{2}(\alpha - \beta)^{2}$